Vector Product (Part-5) | S N Dey

In the previous article, we discussed Short Answer Type Questions of Vector Product. In this article, we will discuss the solutions of Long Answer Type Questions (1-11) of Ex-2A in the chapter Product of Two Vectors as given in the Chhaya Publication Book of aforementioned chapter of S N De book.

Vector Product | S N Dey mathematics class 12 Solutions of Ex-2A

$1.~~$ Applying vectors, show that

$(a_1b_1+a_2b_2+a_3b_3)^2 \leq (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)$

Solution.

let $~~\vec{a}=a_1\hat{i}+a_2\hat{j} +a_3\hat{k},~~\vec{b}=b_1\hat{i}+b_2\hat{j} +b_3\hat{k}$

$\text{We know,}~~~\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}|\cos\theta \\ \text{or,}~~ \vec{a} \cdot \vec{b} \leq|\vec{a}||\vec{b}| ~~~[\because~~\cos\theta \leq 1] \\ \text{or,}~~(a_1\hat{i}+a_2\hat{j} +a_3\hat{k}) \cdot (b_1\hat{i}+b_2\hat{j} +b_3\hat{k}) \\~~\leq \sqrt{a_1^2+a_2^2+a_3^2}\cdot \sqrt{b_1^2+b_2^2+b_3^2} \\ \text{or,}~~a_1b_1+a_2b_2+a_3b_3 \leq \sqrt{a_1^2+a_2^2+a_3^2}\cdot \sqrt{b_1^2+b_2^2+b_3^2} \\ \text{or,}~~ (a_1b_1+a_2b_2+a_3b_3)^2 \leq (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)~~~\text{(showed)}$

$2.~~$ By vector method show that,

$(i)~~$ an angle inscribed in a semi- circle is a right angle.

Solution.

let $~~~\angle ACB~~$ be the inscribed angle .

To prove : $~~\angle ACB =90^{\circ}.$

$~\vec{AB}~~$ is the diameter and $~~O~~$ is the center of the semi-circle.

Now, by the law of triangle of vectors, we get

$\vec{AC}=\vec{AO}+\vec{OC}=\vec{OB}+\vec{OC}~\rightarrow(1) \\~~~~~~~[~\because~ \vec{AO}=\vec{OB}]$

$\vec{OC}+\vec{CB}=\vec{OB} \\ \text{or,} ~~ \vec{CB}=\vec{OB}-\vec{OC} \rightarrow(2)$

Hence, by $~~(1)~~$ and $~~(2)~~$ we get,

$\vec{AC} \cdot \vec{CB}\\=(\vec{OB}+\vec{OC}) \cdot (\vec{OB}-\vec{OC}) \\=|\vec{OB}|^2-|\vec{OC}|^2 \\= 0~~[~\because~~ |\vec{OB}|=|\vec{OC}|=\text{radius}] \\~~\\ ~ \therefore~~ \vec{AC} \cdot \vec{CB}=0 \\ \Rightarrow ~ \vec{AC} \perp \vec{CB}.$

$~\text{Hence,}~~ \angle ACB =90^{\circ}~~~\text{(proved)}$

$(ii)~~$ the parallelogram whose diagonals are equal is a rectangle.

Solution.

let $~~ABCD~~$ be a parallelogram and $~~|\vec{AC}|=|\vec{BD}|.$

$\therefore~|\vec{AC}|^2=|\vec{BD}|^2 \\ \text{or,}~~ |\vec{AB}+\vec{BC}|^2=|\vec{BC}+\vec{CD}|^2 \\ \text{or,}~~ |\vec{AB}+\vec{BC}|^2=|\vec{BC}-\vec{AB}|^2 ~~[\because~~\vec{CD}=-\vec{AB}] \\ \text{or,}~~ |\vec{AB}|^2+|\vec{BC}|^2+2\vec{AB}\cdot \vec{BC}=|\vec{BC}|^2+|\vec{AB}|^2-2\vec{BC}\cdot \vec{AB} \\ \text{or,}~~ 2\vec{AB}\cdot \vec{BC}=-2\vec{AB}\cdot \vec{BC} \\ \text{or,}~~ 4\vec{AB}\cdot \vec{BC}=0 \\ \text{or,}~~ \vec{AB}\cdot \vec{BC}=0\\ \text{So,}~~\vec{AB} \perp \vec{BC}$

Hence, $~~ABCD~~$ is a parallelogram.

$(iii)~~$ the perpendicular bisectors of the sides of a triangle are concurrent.

Solution.

let $~~\vec{OD}~~\text{and}~~\vec{OE}~~$ be the perpendicular bisectors of the sides $~~\vec{BC}~~\text{and}~~ \vec{AC}~~$ respectively, where $~~O~~$ be the point of intersection of $~~\vec{OD}~~\text{and}~~\vec{OE}.$

Suppose that position vectors of $~~A,~B,~C~~$ with respect to $~~O~~$ are $~~\vec{a},~\vec{b},~\vec{c}~~$ respectively.

Let $~~F~~$ be the midpoint of $~~ \vec{AB}~~$ and so the position vectors of $~~D,E,F~~$ are $~~\frac{\vec{b}+\vec{c}}{2},~\frac{\vec{a}+\vec{c}}{2},~~\frac{\vec{a}+\vec{b}}{2}~~$ respectively.

We have to prove that $~~ \vec{OF} \perp \vec{AB}.$

$\text{Now,}~~ \vec{OD} \perp \vec{BC} \\ \therefore~~ \frac{\vec{b}+\vec{c}}{2} \cdot (\vec{c}-\vec{b})=0 \\ \text{or,}~~ (\vec{c}+\vec{b}) \cdot (\vec{c}-\vec{b})=0 \\ \text{or,}~~ |\vec{c}|^2-|\vec{b}|^2=0 \\ \text{or,}~~ |\vec{c}|^2=|\vec{b}| \rightarrow (1)$

$\text{Again,}~~ \vec{OE} \perp \vec{AC} \\ \therefore~~ \frac{\vec{a}+\vec{c}}{2} \cdot (\vec{c}-\vec{a})=0 \\ \text{or,}~~ (\vec{c}+\vec{a}) \cdot (\vec{c}-\vec{a})=0 \\ \text{or,}~~ |\vec{c}|^2-|\vec{a}|^2=0 \\ \text{or,}~~ |\vec{c}|^2=|\vec{a}|^2 \rightarrow (1)$

Hence, from $~~(1)~~$ and $~~(2)~~$ we get,

$~|\vec{a}|^2=|\vec{b}|^2 \\ \therefore~~ |\vec{b}|^2-|\vec{a}|^2=0 \\ \text{or,}~~ (\vec{b}+\vec{a}) \cdot (\vec{b}-\vec{a}) =0 \\ \text{or,}~~ \frac{\vec{a}+\vec{b}}{2}\cdot (\vec{b}-\vec{a})=0\\~~ \\ \therefore~~ \vec{OF} \cdot \vec{AB}=0 \\ \text{So,}~~\vec{OF} \perp \vec{AB}.$

Hence, the perpendicular bisectors of the sides of a triangle are concurrent.

$(iv)~~$ medium to the base of an isoscales triangle is perpendicular to the base.

Solution.

let $~~\Delta ABC~~$ is an isoscales triangle where $~~|\vec{AB}|=|\vec{AC}|\rightarrow(1).$

Suppose that the position vectors of $~~B~~\text{and}~~C~~$ with respect to $~A~$ are $~~\vec{b}~~\text{and}~~\vec{c}~~$ respectively, $~~A~~$ being the origin.

$~\text{So,}~~ \vec{AB}=\vec{b},~~~\vec{AC}=\vec{c}.$

let $~~D~~$ be the mid point of the base $~~BC~~$ so that $~~\vec{AD}=\frac{\vec{b}+\vec{c}}{2}.$

Also, by $~~(1)~~$ we get, $~~|\vec{b}|=|\vec{c}|\rightarrow(2)$

$\therefore~~ \vec{BC} \cdot \vec{AD} \\~~\\~~~~= (\vec{c}-\vec{b}) \cdot \left(\frac{\vec{b}+\vec{c}}{2}\right) \\~~~~= \frac 12[(\vec{c}-\vec{b}) \cdot (\vec{c}+\vec{b})] \\~~~~= \frac 12[|\vec{c}|^2-|\vec{b}|^2] \\~~~~= \frac 12[|\vec{c}|^2-|\vec{c}|^2]~~[\text{By (2)}]\\~~~~= 0 \\~~ \\ \text{So,}~~ \vec{BC} \perp \vec{AD}.$

Hence follows the result.

$3.~~~A(3,-1,2),~~ B(2,-3,3)~~\text{and}~~ C(1,-2,1)~~$ are three given points. Find the angle between the vectors $~~\vec{BA}~~\text{and}~~\vec{BC}.$

Solution.

By the question, the position vectors of $~~A,~B,~C~~$ are given by $~~(3\hat{i}-\hat{j} +2\hat{k}),~~(2\hat{i}-3\hat{j} +3\hat{k}),~~(\hat{i}-2\hat{j} +\hat{k}).$

$\vec{BA}\\= (3\hat{i}-\hat{j} +2\hat{k})-(2\hat{i}-3\hat{j} +3\hat{k}) \\= \hat{i}+2\hat{j} -\hat{k}, \\~~\\~\vec{BC} \\= (\hat{i}-2\hat{j} +\hat{k})-(2\hat{i}-3\hat{j} +3\hat{k}) \\ =-\hat{i}+\hat{j} -2\hat{k}.$

$\therefore~~|\vec{BA}|=\sqrt{1^2+2^2+(-1)^2}=\sqrt{6},\\~~\\~~~~~ |\vec{BC}|=\sqrt{(-1)^2+1^2+(-2)^2}=\sqrt{6}.$

$\text{Again,}~~ \vec{BA} \cdot \vec{BC}\\=(\hat{i}+2\hat{j} -\hat{k}) \cdot (-\hat{i}+\hat{j} -2\hat{k})\\=-1+2+2\\=3$

If $~~\theta~~$ is the angle between the vectors $~~\vec{BA}~~ \text{and}~~\vec{BC},$

$\cos\theta=\frac{\vec{BA} \cdot \vec{BC}}{|\vec{BA}||\vec{BC}|}=\frac{3}{\sqrt{6} \sqrt{6}}\\~~ \\ ~~\therefore~~ \cos\theta=\frac 36=\frac 12 \\ \text{or,}~~\cos\theta=\cos(\pi/3) \\ \text{or,}~~ \theta=\frac{\pi}{3}~~\text{(ans.)}$

$4.~~$ Three vectors $~~\vec{a},~\vec{b},~\vec{c}~~$ are such that $~~\vec{a}+\vec{b}+\vec{c}=\vec{0}~;~$ if $~~ |\vec{a}|=1,~|\vec{b}|=4~~$ and $~~|\vec{c}|=2,~~$ then find the value of $~~\mu=(\vec{a}\cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}).$

Solution.

$~ \vec{a}+\vec{b}+\vec{c}=\vec{0} \\ \text{or,}~~ (\vec{a}+\vec{b}+\vec{c}) \cdot (\vec{a}+\vec{b}+\vec{c}) =\vec{0} \cdot \vec{0} \\ \text{or,}~~ \vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{a}\cdot \vec{c}+\vec{b}\cdot \vec{a}+\vec{b}\cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c}\cdot \vec{a} \\~~ +\vec{c}\cdot \vec{b}+\vec{c}\cdot \vec{c}=0 \\ \text{or,}~~ |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})=0 \\ \text{or,}~~ 1^2+4^2+2^2+2\mu=0 \\ \text{or,}~~ 2\mu=-1-16-4 \\ \text{or,}~~ 2\mu=-21 \\ \text{or,}~~ \mu=-\frac{21}{2}~~~\text{(ans.)}$

$5.~~$ The scalar product of the vector $~~\hat{i}+\hat{j} +\hat{k}~~$ with the unit vector along the sum of vectors $~~2\hat{i}+4\hat{j} -5\hat{k}~~\text{and}~~ \lambda\hat{i}+2\hat{j} +3\hat{k}~~$ is equal to one. Find the value of $~~\lambda.$

Solution.

$\text{let}~ \vec{a}=\hat{i}+\hat{j} +\hat{k},~~\vec{b}=2\hat{i}+4\hat{j} -5\hat{k},\\~~ \vec{c}=\lambda\hat{i}+2\hat{j} +3\hat{k}.$

$\vec{b}+\vec{c}\\=(2\hat{i}+4\hat{j} -5\hat{k})+(\lambda\hat{i}+2\hat{j} +3\hat{k})\\=(2+\lambda)\hat{i}+6\hat{j} -2\hat{k}$

Now, the unit vector along the sum of vectors $~~\vec{b},~~\vec{c}~~$ is given by :

$\hat{n}=\frac{\vec{b}+\vec{c}}{|\vec{b}+\vec{c}|}=\frac{(2+\lambda)\hat{i}+6\hat{j} -2\hat{k}}{\sqrt{(2+\lambda)^2+6^2+(-2)^2}}$

By question,

$\vec{a} \cdot \hat{n}=1 \\ \text{or,}~~ (\hat{i}+\hat{j} +\hat{k}) \cdot \frac{(2+\lambda)\hat{i}+6\hat{j} -2\hat{k}}{\sqrt{(2+\lambda)^2+6^2+(-2)^2}}=1 \\ \text{or,}~~ \frac{(2+\lambda)+6-2}{\sqrt{(2+\lambda)^2+6^2+(-2)^2}}=1 \\ \text{or,}~~ (2+\lambda)+4=\sqrt{(2+\lambda)^2+6^2+(-2)^2} \\ \text{or,}~~ (\lambda+6)^2=(2+\lambda)^2+36+4 \\ \text{or,}~~ \lambda^2+2\lambda \times 6+6^2=2^2+2 \times 2 \times \lambda+\lambda^2+40 \\ \text{or,}~~12 \lambda+36=4+4\lambda+40 \\ \text{or,}~~12\lambda-4\lambda=44-36 \\ \text{or,}~~ 8\lambda=8 \\ \text{or,}~~\lambda=\frac 88=1 ~~\text{(ans.)}$

$6.~~$ Let $~~\vec{a}=2\hat{i}+2\hat{j} +3\hat{k}~,~~\vec{b}=-\hat{i}+2\hat{j} +\hat{k}~~$ and $~~ \vec{c}=3\hat{i}+\hat{j}~~$ be three given vectors ; If $~~\vec{a}+\lambda \vec{b}~~$ and $~~\vec{c}~~$ are perpendicular to each other , find $~~\lambda.$

Solution.

$\vec{a}+\lambda \vec{b}\\=2\hat{i}+2\hat{j} +3\hat{k}+\lambda(-\hat{i}+2\hat{j} +\hat{k})\\=(2-\lambda)\hat{i}+(2+2\lambda)\hat{j}+(3+\lambda)\hat{k} \rightarrow(1)$

$\vec{c}=3\hat{i}+\hat{j}~~~\text{(given)} \rightarrow(2)$

If $~~\vec{a}+\lambda \vec{b}~~$ and $~~\vec{c}~~$ are perpendicular to each other , then

$(\vec{a}+\lambda \vec{b}) \cdot \vec{c}=0 \\ \text{or,}~~ 3(2-\lambda)+(2+2\lambda)+0=0 ~~[\text{By (1),(2)} ] \\ \text{or,}~~ 6-3\lambda+2+2\lambda=0 \\ \text{or,}~~ 8-\lambda=0 \\ \text{or,}~~ \lambda=8 ~~~~ \text{(ans.)}$

$7(i)~~$ Let $~~ \vec{a}=\hat{i}+4\hat{j} +2\hat{k},~~\vec{b}=3\hat{i}-2\hat{j} +7\hat{k}~~$ and $~~\vec{c}=2\hat{i}-\hat{j} +4\hat{k}.~~$ Find a vector $~~\vec{d}~~$ which is perpendicular to both $~~\vec{a}~~\text{and}~~ \vec{a}~~~\text{and}~~\vec{c} \cdot \vec{d}=18.$

Solution.

We know that $~~\vec{a} \times \vec{b}~~$ denotes a vector which is perpendicular to both $~~\vec{a}~~$ and $~~\vec{b}.$

$\vec{a} \times \vec{b} \\= \begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 1& 4 & 2 \\ 3& -2 &7 \\ \end{vmatrix} \\= (28+4)\hat{i}-(7-6)\hat{j} +(-2-12)\hat{k}\\=32\hat{i}-\hat{j}-14\hat{k}\rightarrow(1)$

By the question and using $~~(1),~~$ we can say that

$\vec{d}=\lambda(32\hat{i}-\hat{j}-14\hat{k}) ~~$ where $~~\lambda (\neq 0)~~$ is any scalar, so that

$\vec{c} \cdot \vec{d}=18 \\ \text{or,}~~ (2\hat{i}-\hat{j} +4\hat{k}) \cdot \lambda(32\hat{i}-\hat{j}-14\hat{k})=18 \\ \text{or,}~~ \lambda(64+1-56)=18 \\ \text{or,}~~ 9\lambda=18 \\ \therefore~~\lambda=\frac{18}{9}=2 \rightarrow(2)$

$~\therefore~~\vec{d}\\=\lambda(32\hat{i}-\hat{j}-14\hat{k})\\=2(32\hat{i}-\hat{j}-14\hat{k})~~[\text{By (2)}]\\=64\hat{i}-2\hat{j}-28\hat{k} ~~\text{(ans.)}$

$(ii)~~$ If $~~\hat{i}+\hat{j} +\hat{k} ,~~2\hat{i}+5\hat{j} ,~~3\hat{i}+2\hat{j} -3\hat{k}~~$ and $~~\hat{i}-6\hat{j} -\hat{k}~~$ are the position vectors of points $~~A,B,C~~$ and $~~D~~$ respectively, then find the angle between $~~\overline{AB}~~$ and $~~\overline{CD}.~~$ Deduce that $~~\overline{AB}~~$ and $~~\overline{CD}~~$ are collinear.

Solution.

$\overline{AB}\\=(2\hat{i}+5\hat{j} )-(\hat{i}+\hat{j} +\hat{k}) \\=\hat{i}+4\hat{j} -\hat{k}, ~~\\ \overline{CD}\\= (\hat{i}-6\hat{j} -\hat{k})-(3\hat{i}+2\hat{j} -3\hat{k}) \\= -2\hat{i}-8\hat{j} +2\hat{k}. \\~~\\~~ |\overline{AB}|=\sqrt{1^2+4^2+(-1)^2}=\sqrt{18} \\~~ |\overline{CD}|=\sqrt{(-2)^2+(-8)^2+2^2}=\sqrt{72}.$

$\text{Also,}~~ \overline{AB} \cdot \overline{CD}\\= (\hat{i}+4\hat{j} -\hat{k}) \cdot (-2\hat{i}-8\hat{j} +2\hat{k}) \\=-2-32-2 \\=-36.$

If $~~\theta~~$ be the angle between $~~\overline{AB}~~$ and $~~\overline{CD},~~$ then

$\cos\theta=\frac{\overline{AB} \cdot \overline{CD}}{|\overline{AB}||\overline{CD}|}=\frac{-36}{\sqrt{18} \sqrt{72}} \\ \text{or,}~~ \cos\theta=-\frac{36}{\sqrt{18 \times 18 \times 4}}=-\frac{36}{18 \times 2} \\ \therefore~ \cos\theta=-1=\cos\pi \\ \text{or,}~~ \theta=\pi.$

$\text{Moreover,}~~ \overline{CD}\\=-2\hat{i}-8\hat{j} +2\hat{k} \\=-2(\hat{i}+4\hat{j} -\hat{k})\\=-2\overline{AB}$

Hence, we can deduce that $~~\overline{AB}~~$ and $~~\overline{CD}~~$ are collinear.

$(iii)~~$ Express the vector $~~\vec{a}=5\hat{i}-2\hat{j} +5\hat{k}~~$ as sum of two vectors such that one is parallel to the vector $~~\vec{b}=3\hat{i}+\hat{k} ~~$ and the other is perpendicular to $~~\vec{b}.$

Solution.

Let $~~\vec{a}=\vec{p}+\vec{q}$ where $~~\vec{p}~~$ is parallel to $~~\vec{b}~~$ and $~~\vec{q} \perp \vec{b}.$

Since $~~\vec{p}~~$ is parallel to $~~\vec{b}~~$ so that $~~\vec{p}=\lambda \vec{b}=\lambda(3\hat{i}+\hat{k})\rightarrow(2)~~$ , where $~~\lambda~~$ is a non-zero scalar.

$\text{Now,}~~ \vec{a}=\vec{p}+\vec{q} \\ \text{or,}~~ \vec{q}=\vec{a}-\vec{p} \\ \text{or,}~~ \vec{q}=5\hat{i}-2\hat{j} +5\hat{k}-\lambda(3\hat{i}+\hat{k}) \\ \text{or,}~~ \vec{q}= (5-3\lambda) \hat{i}-2\hat{j}+(5-\lambda)\hat{k} \rightarrow(2)$

Since $~~\vec{q} \perp \vec{b},$

$\vec{q} \cdot \vec{b}=0 \\ \text{or,}~~ 3(5-3\lambda)+0(-2)+1(5-\lambda)=0 \\ \text{or,}~~ 15-9\lambda+5-\lambda=0 \\ \text{or,} ~~ 20-10\lambda=0 \\ \text{or,}~~ 20=10\lambda \\ \therefore~ \lambda =\frac{20}{10}=2$

Hence, by $~~(1)~~\text{and}~~(2),~~$ we get by putting the value of $~~\lambda=2,$

$\vec{p}=2(3\hat{i}+\hat{k}),\\~~\vec{q}=(5-3\times 2) \hat{i}-2\hat{j}+(5-2)\hat{k} =-\hat{i}-2\hat{j}+3\hat{k}.$

$\therefore~ \vec{a}=5\hat{i}-2\hat{j} +5\hat{k}=2(3\hat{i}+\hat{k})+(-\hat{i}-2\hat{j}+3\hat{k}).$

$8.$ If $\vec{a}=\vec{0}\text{or,} \vec{b}=\vec{0},$ then $\vec{a} \cdot \vec{b}=0;~~$ show by an example that the converse of this statement is not always true.

Solution.

Let $~~\vec{a}=2\hat{i}+\hat{j} +\hat{k},~~\vec{b}=\hat{i}+\hat{j} -3\hat{k}.$

Here, clearly $~~\vec{a} \neq 0,~~\vec{b} \neq 0~~$ but

$\vec{a} \cdot \vec{b}\\=(2\hat{i}+\hat{j} +\hat{k}) \cdot (\hat{i}+\hat{j} -3\hat{k})\\=2+1-3\\=0\rightarrow(1)$

Hence by $~(1),~~$ the result follows.

$9.~~$ If $~~\vec{a}=3\hat{i}+4\hat{j} -\hat{k},~~\vec{b}=\hat{i}-3\hat{j} +4\hat{k}~~$ and $~~\vec{c}=5\hat{i}-6\hat{j} +4\hat{k},~~$ find the vector $~~\vec{r}~~$ which is perpendicular to both $~~\vec{a}~~$ and $~~\vec{b}~~$ and which satisfies the relation $~~\vec{r} \cdot \vec{c}=91.$

Solution.

let $~~\vec{r}=x\hat{i}+y\hat{j} +z\hat{k} \rightarrow(1)$

By question,

$\vec{r} \cdot \vec{a}=0 ,~~\vec{r} \cdot \vec{b}=0$

$\vec{r} \cdot \vec{a}=0 \\ \text{or,}~~ (x\hat{i}+y\hat{j} +z\hat{k}) \cdot (3\hat{i}+4\hat{j} -\hat{k})=0 \\ \text{or,}~~ 3x+4y-z=0\rightarrow(2)$

$\vec{r} \cdot \vec{b}=0 \\ \text{or,}~~ (x\hat{i}+y\hat{j} +z\hat{k})\cdot (\hat{i}-3\hat{j} +4\hat{k})=0 \\ \text{or,}~~ x-3y+4z=0 \rightarrow(3)$

From $~~(2)~~$ and $~~(3)~~$ we get by cross multiplication,

$\frac{x}{16-3}=\frac{y}{-1-12}=\frac{1}{-9-4} \\ \text{or,}~~ \frac{x}{13}=\frac{y}{-13}=\frac{z}{-13}=\lambda~ ~[\neq 0~~ \text{(say)}] \\~~\\ \therefore~~ x= 13 \lambda,~~y=-13\lambda,~~z=-13\lambda ~\rightarrow(4)$

$\text{Now,}~~ \vec{r} \cdot \vec{c}=91 \\ \text{or,}~~ (x\hat{i}+y\hat{j} +z\hat{k}) \cdot (5\hat{i}-6\hat{j} +4\hat{k})=91 \\ \text{or,}~~ 5x-6y+4z=91 \\ \text{or,}~~ 13\lambda(5+6-4)=91 \\ \text{or,}~~ 13 \lambda \times 7=91 \\ \text{or,}~~ 91\lambda=91 \\ \therefore~~\lambda =\frac{91}{91}=1\rightarrow(5)$

$~\text{So,}~~ x=13,~~y=-13,~~z=-13~~~\text{(ans.)}$

$10.~~$ Let $~~vec{a},~~\vec{b},~~\vec{c}~~$ be the position vectors of the vertices of a triangle ; prove that the area of the triangle is $~~\frac 12|\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}|.$

Solution.

Area of $~~\Delta ABC~~$ is given by :

$=\frac 12|\vec{AB} \times \vec{BC}|\\~~~~=\frac 12|(\vec{b}-\vec{a}) \times (\vec{c}-\vec{b})|\\~~~~=\frac 12|\vec{b} \times \vec{c}-\vec{b} \times \vec{b}-\vec{a} \times \vec{c}+\vec{a} \times \vec{b}|\\~~~~=\frac 12|\vec{b} \times \vec{c}-0+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}|\\~~~~=\frac 12|\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}|~~\text{(proved)}$

$11.~~$ Given $~~ \vec{a}=4\hat{i}+5\hat{j} -\hat{k},~~\vec{b}=\hat{i}-4\hat{j} +5\hat{k}.~~$ If $~~|\vec{c}|=21~~$ and $~~\vec{c}~~$ is perpendicular to $~~\vec{a}~~$ and $~~\vec{b}~~$ find in component form the vector $~~\vec{c}.$

Solution.

We first compute the value of $~~\vec{a} \times \vec{b}.$

$\vec{a} \times \vec{b} \\= \begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 4 & 5 & -1 \\ 1& -4 & 5 \\ \end{vmatrix} \\ = (25-4)\hat{i}-(20+1)\hat{j} +(-16-5)\hat{k} \\= 21\hat{i}-21\hat{j} -21\hat{k} \\~~\\~~\therefore~ |\vec{a} \times \vec{b}|\\=\sqrt{21^2+(-21)^2+(-21)^2}\\=\sqrt{21^2(1+1+1)}\\=21\sqrt{3}.$

If $~~\vec{c}~~$ is perpendicular to $~~\vec{a}~~$ and $~~\vec{b},~~$ then

$\vec{c}=\lambda(\vec{a} \times \vec{b}), ~~[~~\lambda \neq 0] \rightarrow(1)\\ ~\therefore~ |\vec{c}|=|\lambda| |\vec{a} \times \vec{b}| \\ \text{or,}~~ 21=|\lambda| \times 21\sqrt{3} \\ \text{or,}~~ |\lambda|=\frac{21}{21\sqrt{3}} \\ \text{or,}~~ |\lambda|=\frac{1}{\sqrt{3}} \\ \therefore~ \lambda=\pm \frac{1}{\sqrt{3}}.$

Hence, by $~~(1)~~$ we get

$\vec{c}=\pm\frac{1}{\sqrt{3}}(21\hat{i}-21\hat{j} -21\hat{k}) \\ \text{or,}~~\vec{c}=\pm\frac{21}{\sqrt{3}}(\hat{i}-\hat{j} -\hat{k})~~\text{(ans.)}$

Post Views: 186

Vector Product (Part-5) | S N Dey | Class 12

Vector Product | S N Dey mathematics class 12 Solutions of Ex-2A

$(ii)~~$ the parallelogram whose diagonals are equal is a rectangle.

$(iii)~~$ the perpendicular bisectors of the sides of a triangle are concurrent.

$8.$ If $\vec{a}=\vec{0}\text{or,} \vec{b}=\vec{0},$ then $\vec{a} \cdot \vec{b}=0;~~$ show by an example that the converse of this statement is not always true.

Leave a Comment Cancel reply

Vector Product | S N Dey mathematics class 12 Solutions of Ex-2A

the parallelogram whose diagonals are equal is a rectangle.

the perpendicular bisectors of the sides of a triangle are concurrent.

If then show by an example that the converse of this statement is not always true.

Leave a Comment Cancel reply

$(ii)~~$ the parallelogram whose diagonals are equal is a rectangle.

$(iii)~~$ the perpendicular bisectors of the sides of a triangle are concurrent.

$8.$ If $\vec{a}=\vec{0}\text{or,} \vec{b}=\vec{0},$ then $\vec{a} \cdot \vec{b}=0;~~$ show by an example that the converse of this statement is not always true.